TITLE: Converting sums of square-roots to nested square-roots
QUESTION [45 upvotes]: When solving different equations, I have realised, that some roots containing only arithmetic operations and square roots (4th, 8th roots too, because they can be represented using only square roots) can be converted to nested square roots form. Examples (these are roots of equations of 2nd, 4th, 4th and 8th degree):
$$\sqrt{2}+\sqrt{3}=\sqrt{5+\sqrt{24}}$$
$$\sqrt{2}+\sqrt{3}+\sqrt{6}=\sqrt{15+\sqrt{160+\sqrt{6912+\sqrt{18874368}}}}$$
$$1+\sqrt{2}+\sqrt{3}+\sqrt{6}=\sqrt{21+\sqrt{413+\sqrt{4656+\sqrt{16588800}}}}$$
$$\sqrt{2}+\sqrt{3}+\sqrt{5}=\sqrt{14+\sqrt{140+\sqrt{4096+\sqrt{8847360}}}}$$
However, I have failed to convert to such form following root (8th degree equation):
$$3+\sqrt{2}+\sqrt{3}+\sqrt{5}$$
Performing any operations with it, number of square roots inside increases, what makes me think that converting that root is impossible.
So, question: Can it be done with that root and with what roots in general?
Some forms I was able to get:
$$\sqrt{19+6 \sqrt{2}+6 \sqrt{3}+6 \sqrt{5}+2 \sqrt{6}+2 \sqrt{10}+2 \sqrt{15}}$$
$$\sqrt{19+2\left(\sqrt{33+6 \sqrt{30}}+\sqrt{37+6 \sqrt{30}}+\sqrt{51+6 \sqrt{30}}\right)}$$
If one don't know how I got those expressions, here you are an example.
$$\sqrt{2}+\sqrt{3}+\sqrt{5}=\sqrt{\left(\sqrt{2}+\sqrt{3}+\sqrt{5}\right)^2}=$$
$$=\sqrt{10+2 \left(\sqrt{15}+\sqrt{6}+\sqrt{10}\right)}=\sqrt{10+2 \left(\sqrt{15}+\sqrt{\left(\sqrt{6}+\sqrt{10}\right)^2}\right)}=$$
$$=\sqrt{10+2 \left(\sqrt{15}+\sqrt{16+4 \sqrt{15}}\right)}=\sqrt{10+2 \left(\sqrt{15}-a+a+\sqrt{16+4 \sqrt{15}}\right)}=$$
$$=\sqrt{10+2a+2 \left(\sqrt{\left(\sqrt{15}-a\right)^2}+\sqrt{16+4 \sqrt{15}}\right)}=$$
$$=\sqrt{10+2a+2 \left(\sqrt{15+a^2-2a \sqrt{15}}+\sqrt{16+4 \sqrt{15}}\right)}=$$
$$[2a=4 \Rightarrow a=2]$$
$$=\sqrt{14+2 \left(\sqrt{19-4\sqrt{15}}+\sqrt{16+4 \sqrt{15}}\right)}=$$
$$=\sqrt{14+2 \sqrt{\left(\sqrt{19-4\sqrt{15}}+\sqrt{16+4 \sqrt{15}}\right)^2}}=$$
$$=\sqrt{14+2 \sqrt{35+4 \sqrt{16+3 \sqrt{15}}}}=\sqrt{14+\sqrt{140+\sqrt{4096+\sqrt{8847360}}}}$$

REPLY [3 votes]: Here is other approach. I hope it is most helpful.
A.

For any vector $v$ with integer coefficients
$$
v = (a,b,c,d,e,f,g,h)
$$
denote linear combination
$$
x_v = a+b\sqrt{2}+c\sqrt{3}+d\sqrt{5}+e\sqrt{6}+f\sqrt{10}+g\sqrt{15}+h\sqrt{30}.\tag{1}
$$
Values $x_v$ generate ring (all sums, products of them has form $(1)$ etc).
Value
$$
x_v^2-A
$$
has according/appropriate vector with integer coefficients (vector $w=(a',b',c',d',e',f',g',h')$, where its coefficients are defined as
$$
\begin{array}{l}
a'=a^2+2b^2+3c^2+5d^2+6e^2+10f^2+15g^2+30h^2-A;\\
b'=2(ab+3ce+5df+15gh),\\
c'=2(ac+2be+5dg+10fh),\\
d'=2(ad+2bf+3cg+6eh),\\
e'=2(ae+bc+5fg+5dh),\\
f'=2(af+bd+3eg+3ch),\\
g'=2(ag+cd+2ef+2bh),\\
h'=2(ah+bg+cf+de).\tag{2}
\end{array}
$$
So, one can define vector transformation
$$
w = T(v,A),\tag{3}
$$
described by $(2)$.

B.

If there is representation
$$
3+\sqrt{2}+\sqrt{3}+\sqrt{5}=\sqrt{A_1+\sqrt{A_2+\sqrt{A_3+...+\sqrt{A_{n-1}+\sqrt{A_n}}}}},\tag{4}
$$
then for vector $v_0=(3,1,1,1,0,0,0,0)$  exist these transformations:
$$
v_1=T(v_0,A_1),\\
v_2=T(v_1,A_2),\\
...\\
v_n=T(v_{n-1},A_n),\tag{5}
$$
where $v_n$ iz zero-vector: $v_n=(0,0,0,0,0,0,0,0)$.

C.
Now, apply modular arithmetic to $(2), (3), (5)$.
For $m\in\mathbb{N}$ denote transformation
$$
z = T_m(v,A),\tag{3'}
$$
where $z=(a'\bmod m, ~ b'\bmod m, ~ c'\bmod m, ... ,  ~ h'\bmod m)$,
and $a',b',c',...,h'$ are defined in $(2)$.
If there exist transformations $(5)$  converting $v_0=(3,1,1,1,0,0,0,0)$ to $v_n=(0,0,...,0)$, then for any $m\in\mathbb{N}$ must exist transformations
$$
z_1=T_m(v_0,A_1),\\
z_2=T_m(z_1,A_2),\\
...\\
z_n=T_m(z_{n-1},A_n),\tag{5'}
$$
where $z_n=(0,0,0,0,0,0,0,0)$.

D.
For $n=4$ ($4$ nested rdicals), if we choose $m=5,7,10,11,...$, then for $0\le A_1,A_2,A_3,A_4 <m$ value $z_4$ cannot be zero-vector. 
For $n=5$ one can use $m=11,13,17,19,...$ to see that $z_5$ cannot be $\vec{0}$.
For $n=6$ one can use $m=31, 41, 43, 47, ...$ to see that $z_6$ cannot be $\vec{0}$.
So, must be at least $n\ge 7$ radicals.
(I'll update this answer after checking $n=7$).

How D is working (temporary part):
Example for $n=5$, $m=11$:
starting vector is (a[0]=3,b[0]=1,c[0]=1,d[0]=1,e[0]=...=h[0]=0)
I need to check if there is sequence $A[0],...,A[4]$ that generates (0,0,0,0,0,0,0,0).
It can be used "as is", or more smart:
if there is sequence $A[0], ..., A[2],A[3]$ that generates vector (a,b,c,d,e,f,g,h),
where $a$ can be nonzero;
or even so:
if there is sequence $A[0], ..., A[2]$ that generates vector (a,b,c,d,e,f,g,h),
where $a$ can be nonzero and one of $b,c,d,e,f,g,h$ is nonzero too.
If apply some modulo $m$, then each  $A[j]$ is equivalent to $0$ or $1$ or $2$ or ... or $m-1$.
Then test:
for A[0]=0,1, ..., m-1
for A[1]=0,1, ..., m-1
for A[2]=0,1, ..., m-1
{
  a[1] =  (a[0]*a[0]+ 30*h[0]*h[0] - A[0]) mod m;
  b[1] = 2(a[0]*b[0]+...+15*g[0]*h[0]) mod m;
  ...
  h[1] = 2(a[0]*h[0]+...+d[0]*e[0]) mod m;

  a[2] =  (a[1]*a[1]+ 30*h[1]*h[1] - A[1]) mod m;
  b[2] = 2(a[1]*b[1]+...+15*g[1]*h[1]) mod m;
  ...
  h[2] = 2(a[1]*h[1]+...+d[1]*e[1]) mod m;

  and 
  a[3], b[3],   ...,  h[3] same way;

  and 
  a[4], b[4],   ...,  h[4] same way, but use 0 instead of A[3];

  then check if only one of b[4],c[4],...,h[4] is nonzero (say d[4]);
  if "yes", then success (possible solution);
  then A[3] = a[4], and A[4] is based on this last non-zero term (like d[4]);   
}